Solved – the probability P(X > Y) given X ~ Be(a1, b1), and Y ~ Be(a2, b2), and X and Y are independent

Given two independent variables with Beta distribution, \$X sim text{Be}(a_1, b_1)\$ and \$Y sim text{Be}(a_2, b_2)\$, how do you find the probability that the value of X is greater than the value of Y for a given observation?

Does this probability have a name that I'm just blanking on?

Contents

\$Pr(X > Y) = int_0^1 frac{x^{a_1 – 1}(1 – x)^{b_1 – 1}}{text{Be}(a_1,b_1)} int_0^xfrac{y^{a_2 – 1}(1 – y)^{b_2 – 1}}{text{Be}(a_2,b_2)} dy dx\$

\$Pr(X > Y) = frac{1}{text{Be}(a_1,b_1)} int_0^1 x^{a_1 – 1}(1 – x)^{b_1 – 1}I_x(a_2, b_2) dx\$

where \$I_x(a, b)\$ is the regularized incomplete beta function. If \$a\$ and \$b\$ are integers then

\$I_x(a,b) = sum_{j=a}^{a+b-1} {(a+b-1)! over j!(a+b-1-j)!} x^j (1-x)^{a+b-1-j}.\$

Substitute in, do some simple algebra, and the integral will have a closed form solution as a finite sum of beta functions.

If \$a_2\$ and \$b_2\$ aren't integers but \$a_1\$ and \$b_1\$ are, then calculate \$Pr(X > Y) = 1 – Pr(Y > X)\$. If neither case holds, you're pooched for an analytical solution but you can always do the integral numerically, either deterministically or by Monte Carlo.

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